2023-I-03

The maximum absolute error correct to the nearest $10\text{ g}$

$\begin{array}{cl}=& {\displaystyle \frac{1}{2}}\times 10\\ =& 5\text{ g}\end{array}$

Therefore, the minimum possible weight of a regular packet of cheese

$\begin{array}{cl}=& 220-5\\ =& 215\text{ g}\end{array}$

Therefore, the minimum possible weight of $250$ regular packets of cheese

$\begin{array}{cl}=& 215\times 250\\ =& 53750\text{ g}\\ =& 53.75\text{ kg}\end{array}$

The maximum absolute error correct to the nearest $0.1\text{ kg}$

$\begin{array}{cl}=& {\displaystyle \frac{1}{2}}\times 0.1\\ =& 0.05\text{ kg}\end{array}$

Therefore, the upper limit of $53.6\text{ kg}$

$\begin{array}{cl}=& 53.6+0.05\\ =& 53.65\text{ kg}\end{array}$

Since the minimum possible weight of $250$ regular packets of cheese is greater than the upper limit of $53.6\text{ kg}$, then the claim is not correct.